The equation of the plane through the point \((2,-1,-3)\) and parallel to the lines \(\frac{x-1}{3}=\frac{y+2}{2}=\frac{z}{-4}\) and \(\frac{x}{2}=\frac{y-1}{-3}=\frac{z-2}{2}\) is

Equation of the plane passing through \(\overline{\mathrm{a}}\) and parallel to \(\overline{\mathrm{b}}\) and \(\overline{\mathrm{c}}\) is \(\overline{\mathrm{r}} \cdot(\overline{\mathrm{b}} \times \overline{\mathrm{c}})=\overline{\mathrm{a}} \cdot(\overline{\mathrm{b}} \times \overline{\mathrm{c}})\)
\(\begin{aligned} & \therefore \quad \overline{\mathrm{b}} \times \overline{\mathrm{c}}=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 3 & 2 & -4 \\ 2 & -3 & 2\end{array}\right|=-8 \hat{\mathrm{i}}-14 \hat{\mathrm{j}}-13 \hat{\mathrm{k}} \\ & \begin{aligned} \therefore & \overline{\mathrm{a}} \cdot(\overline{\mathrm{b}} \times \overline{\mathrm{c}}) \\ & =(2)(-8)+(-1)(-14)+(-3)(-13) \\ & =-16+14+39=37\end{aligned}\end{aligned}\)
\(\therefore \quad\) Required equation is
\(\bar{r} \cdot(-8 \hat{\mathbf{i}}-14 \hat{\mathrm{j}}-13 \hat{\mathrm{k}})=37\)
i.e., \(8 x+14 y+13 z+37=0\)